A brick of mass 2.0kg is at rest. It falls to the ground through a distance of 5.0 m. The acceleration of free fall g is 10 m/s2.
Air resistance can be ignored. At what speed does the brick hit
the ground?

I may not have the answer so i'll just give up some hints.

Multiply the time by the acceleration due to gravity to find the velocity when the object hits the ground. If it takes 9.9 seconds for the object to hit the ground, its velocity is (1.01 s)*(9.8 m/s^2), or 9.9 m/s. Choose how long the object is falling. In this example, we will use the time of 8 seconds. Calculate the final free fall speed (just before hitting the ground) with the formula v = v₀ + gt = 0 + 9.80665 * 8 = 78.45 m/s . Find the free fall distance using the equation s = (1/2)gt² = 0.5 * 9.80665 * 8² = 313.8 m .h = 0.5 * 9.8 * (1.5)^2 = 11m. b. V = gt = 9.8 * 1.5 = 14.7m/s. A feather and brick dropped together. Air resistance causes the feather to fall more slowly. If a feather and a brick were dropped together in a vacuum—that is, an area from which all air has been removed—they would fall at the same rate, and hit the ground at the same time.When an object's point is taller the thing that is going down it will go faster than when the point is lower. EXAMPLE: The object is the tennis ball if you drop it down the higher hill it will be faster than if you drop it down a shorter hill. In other words, if two objects are the same size but one is heavier, the heavier one has greater density than the lighter object. Therefore, when both objects are dropped from the same height and at the same time, the heavier object should hit the ground before the lighter one.

I hope my little bit (big you may say) hint help you with your question.

The total momentum before and after the collision must be conserved.

The total momentum before the collision is:

where m1 and m2 are the masses of the two players, and  and  their initial velocities. Both are considered with positive sign, because the two players are running toward the same direction.

The final momentum is instead

because now the two players are moving together with a total mass of (m1+m2) and final speed vf.

By requiring that the momentum is conserved

we  can calculate vf, the post-collision speed:


and the direction is the same as the direction of the players before the collision.